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Q.
The acceleration due to gravity on the surface of a satellite is $1.5\,ms^{-2}$, the time period of a simple pendulum on it, if its time period on earth is r, will be
Oscillations
Solution:
$On \, earth$ Time period, $T_e = 2\pi \sqrt{\frac{l}{ge}}$ , or $t = 2\pi \sqrt{\frac{1}{9.8}} $ ....(i)
$On \, moon$ $T_p = 2 \pi \sqrt{\frac{l}{g_m}}$ or $T_p = 2 \pi \sqrt{\frac{l}{1.5}}$ .....(ii)
Dividing (ii) by (i), $\frac{T_p}{t} = \sqrt{\frac{9.8}{1.5}} $ or $T_p = t \sqrt{\frac{9.8}{1.5}}$ = 2.56 t